a) 1+2+3+....+ x = 1275 ( vế trái có x số hạng )
( 1 +x ) . x : 2 = 1275
( 1 + x ) . x = 1275 x 2
( 1 + x ) . x = 2550
Vì 2550 = 50 x 51 => x = 50
b, \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)=6138\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+99\right)=6138\)
\(\Rightarrow99\cdot x+4950=6138\)
\(\Rightarrow99\cdot x=6138-4950\)
\(\Leftrightarrow99\cdot x=1188\)
\(\Rightarrow x=\frac{1188}{99}\)
\(\Leftrightarrow x=12\)
a, Ta có công thức tính n : lợi dụng x ta có :
\(1+2+3+4+x=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow\frac{\left(n+1\right)\cdot n}{2}=1275\)
\(\Rightarrow\left(n+1\right)\cdot n=1275\cdot2\)
\(\Rightarrow\left(n+1\right)\cdot n=2550\)
\(\Rightarrow\left(n+1\right)\cdot n=51\cdot50\)
Vậy ta có x = 50
=> x = 50 thỏa mãn
a) \(1+2+3+...+x=1275\)
\(\Leftrightarrow\left(x+1\right).x:2=1275\)
\(\Leftrightarrow x\left(x+1\right)=2550\Leftrightarrow50.51=2550\Rightarrow x=50.\)
b) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)=6138\)
\(\Leftrightarrow99x+\left(1+2+3+...+99\right)=6138\)
\(\Leftrightarrow99x+4950=6138\Leftrightarrow99x=1185\Leftrightarrow x=11,\left(96\right).\)
\(a,\text{ }1+2+3+...+x=1275\)
\(\left[\left(x-1\right)\text{ : }1+1\right]\text{ x }\left(x+1\right)\text{ : }2\)
\(x\text{ x }\left(x+1\right)\text{ : }2=1275\)
\(x\text{ x }\left(x+1\right)=1275\text{ x }2\)
\(x\text{ x }\left(x+1\right)=2550\)
\(\text{Vì }x\text{ x }\left(x+1\right)\text{ là tích của hai số tự nhiên liên tiếp mà }50\text{ x }51=2550\)
\(\Rightarrow\text{ }x=50\)
\(b,\text{ }\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)=6138\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+99\right)=6138\)
\(99x+4950=6138\)
\(99x=6138-4950\)
\(99x=1188\)
\(x=1188\text{ : }99\)
\(x=12\)
